60 lines
1.4 KiB
Python
60 lines
1.4 KiB
Python
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#!/usr/bin/env python3
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# Link: https://crackmes.one/crackme/63c4ee1a33c5d43ab4ecf49a
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# Conditions that must be fulfilled for the key to be valid:
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# key[0] == '%'
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# key[1] << t == 200
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# key[2] << t == 212
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# key[3] == 'k'
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# key[4] << t == 80
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# key[5] & 1073741823 == 57
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# key[6] == '^'
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# key[7] << t == 246
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# key[8] & 1073741823 == 46
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# key[9] == 'f'
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# key[10] << t == 128
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# key[11] & 1073741823 == 49
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# key[12] == 'F'
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# key[13] << t == 104
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# (t = (localtime() % 10) % 3) => t either 0, 1, or 2
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# After these conditions are met the printable string (either "Incorrect password." or
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# "Correct password. Congratulation!") is decoded by shifting each of the string's
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# character's value to the right by one and subtracting 5 from that.
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from string import ascii_letters, digits, punctuation
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import itertools
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key = ["%", "", "", "k", "", "", "^", "", "", "f", "", "", "F", ""]
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for c in itertools.chain(*[ascii_letters, digits, punctuation]):
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shift = ord(c) << 1
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match shift:
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case 200:
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key[1] = c
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case 212:
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key[2] = c
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case 80:
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key[4] = c
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case 246:
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key[7] = c
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case 128:
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key[10] = c
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case 104:
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key[13] = c
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band = ord(c) & 0x3FFFFFFF
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match band:
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case 57:
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key[5] = c
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case 46:
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key[8] = c
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case 49:
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key[11] = c
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print(f"{''.join(key)}")
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